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  3. JEE Main & Advanced

JEE Main & Advanced Sample Paper JEE Main Sample Paper-2

  • question_answer
    A force of 10 N is required to move a conducting loop through a non-uniform magnetic field at a speed 2 m/s.  The rate of production of electrical energy in the loop will be

    A)  100 Watt            

    B)  30 Watt

    C)  zero                     

    D)  20 Watt

    Correct Answer: D

    Solution :

    Rate of loss = Electrical power \[\therefore \] Electrical energy production \[=\vec{F}\cdot \vec{v}=10\times 2=20\]


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